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30 tháng 7 2021
Bài 5 :
f, bạn xem lại đề hay là tìm x chứa tham số a ?
g, \(x^2+3x-\left(2x+6\right)=0\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\Leftrightarrow x=-3;x=2\)
h, \(5x+20-x^2-4x=0\Leftrightarrow5\left(x+4\right)-x\left(x+4\right)=0\)
\(\Leftrightarrow\left(5-x\right)\left(x+4\right)=0\Leftrightarrow x=-4;x=5\)
m, \(x^3-5x^2-x+5=0\Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-5\right)=0\Leftrightarrow x=\pm1;x=5\)
n, \(x\left(x-3\right)-7x+21=0\Leftrightarrow x\left(x-3\right)-7\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-3\right)=0\Leftrightarrow x=3;x=7\)
NT
30 tháng 8 2017
x^2 -6x +10 = x^2 -2.x.3 +3^2 +1 = (x-3)^2 +1
Ma (x-3)^2 >=0 <=> (x-3)^2 +1 >=1>0 (voi moi x)
b) 4x - x^2 -5 = -(x^2 -4x +5) =-[(x^2 -4x +4)+1] = -[(x-2)^2 +1]
Ma (x+2)^2 >=0 <=> (x-2)^2 +1 >=1 <=> -[(x-2)^2 +1] <=-1 => -[(x-2)^2 +1] <0
2) a) P= x^2 -2x +5 = x^2 -2x +1 +4 = (x-1)^2 +4
Ta co: (x-1)^2 >=0 <=> (x-1)^2 +4 >=4
Vay gia tri nho nhat P=4 khi x=1
b) Q= 2x^2 -6x = 2(x^2 -3x) = 2(x^2 - 2.x.3/2 + 9/4 -9/4)= 2[(x-3/2)^2 -9/4]
Ta co: (x-3/2)^2 >=0 <=>(x-3/2)^2 -9/4 >= -9/4 <=> 2[(x-3/2)^2 -9/4] >= -9/2
Vay gia tri nho nhat Q= -9/2 khi x= 3/2
c) M= x^2 +y^2 -x +6y +10 = (x^2 -2.x.1/2 + 1/4) +(y^2 +2.y.3+9)+3/4
= ( x-1/2)^2 + (y+3)^2 +3/4
M>= 3/4
Vay GTNN cua M = 3/4 khi x=1/2 va y=-3
3)a) A= 4x - x^2 +3 = -(x^2 -4x -3) = -( x^2 -4x+4 -7) =-[(x-2)^2 -7]
Ta co: (x-2)^2>=0 <=> (x-2)^2 -7 >=-7 <=> -[(x-2)^2 -7] <=7
Vay GTLN A=7 khi x=2
b) B= x-x^2 = -(x^2 -2.x.1/2+1/4-1/4) = -[(x-1/2)^2 -1/4]
GTLN B= 1/4 khi x=1/2
c) N= 2x - 2x^2 -5 =-2( x^2 -x+5/2) = -2(x^2 - 2.x.1/2 +1/4 +9/4)
= -2[(x-1/2)^2 +9/4]
GTLN N= -9/2 khi x=1/2
A
18 tháng 10 2020
Bài 2 :
a, \(x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow x=0;4\)
b, \(5x\left(x-2020\right)-x+2020=0\)
\(\Leftrightarrow5x\left(x-2020\right)-\left(x-2020\right)=0\Leftrightarrow\left(5x-1\right)\left(x-2020\right)=0\)
\(\Leftrightarrow x=\frac{1}{5};2020\)
c, \(\left(4x+5\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow16x^2+40x+25-\left(4x^2-4x+1\right)=0\)
\(\Leftrightarrow12x^2+44x+24=0\Leftrightarrow4\left(x+3\right)\left(3x+2\right)=0\)
\(\Leftrightarrow x=-3;-\frac{2}{3}\)
30 tháng 7 2021
Bài 5 :
a, \(2x\left(x-3\right)+x-3=0\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)
b, \(x\left(x+1\right)-x-1=0\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=\pm1\)
c, sửa đề \(x^3-3x^2+x-3=0\Leftrightarrow x^2\left(x-3\right)+x-3=0\)
\(\Leftrightarrow\left(x^2+1>0\right)\left(x-3\right)=0\Leftrightarrow x=3\)
d, \(3x^2\left(2x-1\right)+1-4x^2=0\Leftrightarrow3x^2\left(2x-1\right)+\left(1-2x\right)\left(1+2x\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x^2-2x-1\right)=0\Leftrightarrow\left(2x-1\right)\left(3x+1\right)\left(x-1\right)=0\Leftrightarrow x=1;x=-\frac{1}{3};x=\frac{1}{2}\)
e, \(x^3+2x-x^2-2=0\Leftrightarrow x\left(x^2+2\right)-\left(x^2+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+2>0\right)=0\Leftrightarrow x=1\)
13 tháng 7 2021
a, ( 2x - 3 )2- (2x + 1)2 = -3
4x2-12x+9-4x2+4x-1=-3
-8x-1=-3
-8x=-2
x=\(\frac{1}{4}\)
b, (5x - 1) 2 - (5x + 4)(5x - 4) = 7
25x2-10x+1-25x2+16=7
-10x+17=7
-10x=-10
x=1
c, ( x- 5)2 + (x-3)(x+3) - 2(x + 1)2=0
x2-10x+25+x2-9-2x2-4x-2=0
-14x+14=0
-14(x-1)=0
=>x-1=0
x=1
AD
13 tháng 7 2021
a) \(\left(2x-3\right)^2-\left(2x+1\right)^2=-3\)
\(\Leftrightarrow4x^2-12x+9-4x^2-4x-1=-3\)
\(\Leftrightarrow-16x+8=-3\)
\(\Leftrightarrow-16x=-11\)
\(\Leftrightarrow x=\frac{11}{16}\)
b)\(\left(5x-1\right)^2-\left(5x+4\right)\left(5x-4\right)=7\)
\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)
\(\Leftrightarrow-10x+17=7\)
\(\Leftrightarrow-10x=-10\)
\(\Leftrightarrow x=1\)
c)\(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+1\right)^2=0\)
\(\Leftrightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow2x^2-10x-16-2x^2-4x-2=0\)
\(\Leftrightarrow-14x-18=0\)
\(\Leftrightarrow-14x=18\)
\(\Leftrightarrow x=-\frac{9}{7}\)
#H
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
8 tháng 8 2023
a/
\(x^3-4x^2-\left(x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=-1\end{matrix}\right.\)
b/
\(x^5-9x=0\)
\(\Leftrightarrow x\left(x^4-9\right)=x\left(x^2-3\right)\left(x^2+3\right)=0\)
\(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)
c/
\(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)
\(\Leftrightarrow x^4\left(x-1\right)^2-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^4-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2-2\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\pm\sqrt{2}\end{matrix}\right.\)
8 tháng 8 2023
a. x mũ 2 - 2x + 1 = 25
= x^2 + 2.x.1 + 1^2
= ( x + 1 ) ^2
ko bt có đúng ko nữa, mấy câu kia tui ko bt lm
21 tháng 9 2020
a) ( x - 3 )2 - 4 = 0
<=> ( x - 3 )2 - 22 = 0
<=> ( x - 3 - 2 )( x - 3 + 2 ) = 0
<=> ( x - 5 )( x - 1 ) = 0
<=> x = 5 hoặc x = 1
b( 2x + 3 )2 - ( 2x + 1 )( 2x - 1 ) = 22
<=> 4x2 + 12x + 9 - ( 4x2 - 1 ) = 22
<=> 4x2 + 12x + 9 - 4x2 + 1 = 22
<=> 12x + 10 = 22
<=> 12x = 12
<=> x = 1
c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )2 = 16
<=> 16x2 - 9 - ( 16x2 - 40x + 25 ) = 16
<=> 16x2 - 9 - 16x2 + 40x - 25 = 16
<=> 40x - 34 = 16
<=> 40x = 50
<=> x = 50/40 = 5/4
d) x3 - 9x2 + 27x - 27 = -8
<=> ( x - 3 )3 = -8
<=> ( x - 3 )3 = (-2)3
<=> x - 3 = -2
<=> x = 1
e) ( x + 1 )3 - x2( x + 3 ) = 2
<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2
<=> 3x + 1 = 2
<=> 3x = 1
<=> x = 1/3
f) ( x - 2 )3 - x( x - 1 )( x + 1 ) + 6x2 = 5
<=> x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 = 5
<=> x3 + 12x - 8 - x3 + x = 5
<=> 13x - 8 = 5
<=> 13x = 13
<=> x = 1
21 tháng 9 2020
a) \(\left(x-3\right)^2-4=0\)
=> \(\left(x-3\right)^2-2^2=0\)
=> \(\left(x-3-2\right)\left(x-3+2\right)=0\)
=> \(\left(x-5\right)\left(x-1\right)=0\)
=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)
=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)
=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)
=> \(4x^2+12x+9-4x^2+1=22\)
=> \(12x+9+1=22\)
=> \(12x+10=22\)
=> 12x = 12
=> x = 1
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)
=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)
=> \(16x^2-9-16x^2+40x-25=16\)
=> \(-9+40x-25=16\)
=> \(40x=16+25-\left(-9\right)=16+25+9=50\)
=> x = 50/40 = 5/4
d) \(x^3-9x^2+27x-27=-8\)
=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)
=> \(\left(x-3\right)^3=-8\)
=> \(\left(x-3\right)^3=\left(-2\right)^3\)
=> x - 3 = -2 => x = 1
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)
=> \(3x+1=2\)
=> \(3x=1\)=> x = 1/3
f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)
=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)
=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)
=> \(\left(12x+x\right)-8=5\)
=> 13x = 13
=> x = 1
6 tháng 8 2021
a, \(5x\left(x-1\right)+\left(x+17\right)=0\)
\(\Leftrightarrow5x^2-5x+x+17=0\Leftrightarrow5x^2-4x+17=0\)
\(\Leftrightarrow5\left(x^2-\frac{4}{5}x\right)+17=0\Leftrightarrow5\left(x^2-2.\frac{2}{5}x+\frac{4}{25}-\frac{4}{25}\right)+17=0\)
\(\Leftrightarrow5\left(x-\frac{2}{5}\right)^2-\frac{4}{5}+17=0\Leftrightarrow5\left(x-\frac{2}{5}\right)^2+81\ge81>0\)
Vậy pt vô nghiệm
b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)
\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)
\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\Leftrightarrow x.2x=0\Leftrightarrow x=0\)
c, \(2x^2-9x+7=0\Leftrightarrow2x^2-7x-2x+7=0\)
\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\Leftrightarrow\left(x-1\right)\left(2x-7\right)=0\Leftrightarrow x=1;x=\frac{7}{2}\)
QA
6 tháng 8 2021
Trả lời:
a, \(5x\left(x-1\right)+\left(x+17\right)=0\)
\(\Leftrightarrow5x^2-5x+x+17=0\)
\(\Leftrightarrow5x^2-4x+17=0\)
\(\Leftrightarrow5\left(x^2-\frac{4}{5}x+\frac{17}{5}\right)=0\)
\(\Leftrightarrow x^2-\frac{4}{5}x+\frac{17}{5}=0\)
\(\Leftrightarrow x^2-2.x.\frac{2}{5}+\frac{4}{25}+\frac{81}{25}=0\)
\(\Leftrightarrow\left(x-\frac{2}{5}\right)^2+\frac{81}{25}=0\)
Vì \(\left(x-\frac{2}{5}\right)^2+\frac{81}{25}\ge\frac{81}{25}>0\forall x\)
nên pt vô nghiệm
b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)
\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)
\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\)
\(\Leftrightarrow3x.\left(-9\right).2x=0\)
\(\Leftrightarrow-54x^2=0\)
\(\Leftrightarrow x^2=0\)
\(\Leftrightarrow x=0\)
Vậy x = 0 là nghiệm của pt.
c, \(7-9x+2x^2=0\)
\(\Leftrightarrow2x^2-7x-2x+7=0\)
\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=1\end{cases}}}\)
Vậy x = 7/2; x = 1 là nghiệm của pt.
d, trùng ý c
B= 7800 nha
x=3
a=2